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Math Help - Determinant proof

  1. #1
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    Determinant proof

    How do you prove "if det(-A)=0, then det(adj A)=0" ? I am stuck on the right way to prove this. Thank you.
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    Quote Originally Posted by antman View Post
    How do you prove "if det(-A)=0, then det(adj A)=0" ? I am stuck on the right way to prove this. Thank you.
    Hint 1: \det(-A) \implies \det (A) = 0
    Hint 2: A ~\text{adj}(A) = \det (A) I
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    If det(A)=0, then det(-A) is also 0? I do not know what to do with .
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    Quote Originally Posted by antman View Post
    If det(A)=0, then det(-A) is also 0? I do not know what to do with .
    Yes, because \det (-A) = (-1)^n \det (A) = 0. Since A~ \text{adj}(A) = I\det (A) = \bold{0}. If the adjoint was invertible then it would force A = \bold{0} \implies \text{adj}(A) =\bold{0} which is a contradiction. Thus, adjoint is not invertible and so its determinant must be zero.
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    So that proves if det(-A)=0, then det(adj A)=0? and it is indeed true? I am confused about how exactly the adjoint relates to the original matrix, let alone the negative matrix. The adjoint is the transpose of the coefficient matrix, but I am not entirely clear on how to get the coefficient matrix. Would your statement (. Since . If the adjoint was invertible then it would force which is a contradiction. Thus, adjoint is not invertible and so its determinant must be zero.) be a sufficient proof? I am creating an example to go with it. I'm just having a hard time knowing when enough information is given to be considered a proof. Thank you for all of your help!
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    Quote Originally Posted by antman View Post
    be a sufficient proof?
    Yes. I showed that if \text{adj}(A) is invertible then it would form \text{adj}(A) to be the zero matrix, but that is a contradiction. Therefore, \text{adj}(A) is not invertible, but if it is not invertible then the determinant must be zero.
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